The pH of a solution obtained by mixing 5 mL of 0.1 M $$\text{NH}_4\text{OH}$$ solution with 250 mL of 0.1 M $$\text{NH}_4\text{Cl}$$ solution is _____ $$\times 10^{-2}$$. (Nearest integer)
Given: $$\text{pK}_b(\text{NH}_4\text{OH}) = 4.74$$, $$\log 2 = 0.30$$, $$\log 3 = 0.48$$, $$\log 5 = 0.70$$

For a buffer containing a weak base $$\text{B}$$ and its salt $$\text{BH}^+$$, the Henderson-Hasselbalch type relation is

$$\text{pOH} = \text{p}K_b + \log\frac{[\text{salt}]}{[\text{base}]}$$

Step 1 - Calculate moles of each component.

Base ($$NH_4OH)$$: $$0.1\;M \times 5\;{\text{mL}} = 0.1 \times \frac{5}{1000} = 5.0 \times 10^{-4}\;\text{mol}$$

Salt $$(NH_4Cl)$$: $$0.1\;M \times 250\;{\text{mL}} = 0.1 \times \frac{250}{1000} = 2.5 \times 10^{-2}\;\text{mol}$$

Step 2 - Find concentrations after mixing.

Total volume $$V = 5\;{\text{mL}} + 250\;{\text{mL}} = 255\;{\text{mL}} = 0.255\;L$$

$$[\text{NH}_4\text{OH}] = \frac{5.0 \times 10^{-4}}{0.255} \approx 1.96 \times 10^{-3}\;M$$

$$[\text{NH}_4^+] = \frac{2.5 \times 10^{-2}}{0.255} \approx 9.80 \times 10^{-2}\;M$$

Step 3 - Insert into the buffer equation.

Ratio $$\frac{[\text{salt}]}{[\text{base}]} = \frac{9.80 \times 10^{-2}}{1.96 \times 10^{-3}} \approx 50$$

Using the given logarithms:
$$\log 50 = \log(5 \times 10^1) = \log 5 + \log 10 = 0.70 + 1 = 1.70$$

Given $$\text{p}K_b(\text{NH}_4\text{OH}) = 4.74$$, so

$$\text{pOH} = 4.74 + 1.70 = 6.44$$

Step 4 - Find pH.

$$\text{pH} = 14 - \text{pOH} = 14 - 6.44 = 7.56$$

Step 5 - Express in the required form.

$$7.56 = 756 \times 10^{-2}$$
Nearest integer = 756.

Answer: 756